Małgorzata Filipczak, Tomasz Filipczak
On the comparison of the density type topologies generated by sequences and by functions
Abstract. In the paper we investigate density type topologies generated by functions f satisfying condition lim inf
x→0+ f(x)
x
> 0, which are not generated by any sequence.
2000 Mathematics Subject Classification: 54A10, 26A15, 28A05.
Key words and phrases: density points, density topology, comparison of topologies.
Through the paper we shall use standard notation: ℝ will be the set of real numbers, ℕ the set of positive integers, L the family of Lebesgue measurable subsets of ℝ and |E| the Lebesgue measure of a measurable set E. By Φ d (E) we shall denote the set of all Lebesgue density points of measurable set E (i.e. Φ d (E) = n x ∈ ℝ; lim h →0
+|(x−h;x+h)∩E|
2h = 1 o
) and by T d the density topology consists of measurable sets satisfying E ⊂ Φ d (E). For any operators Φ, Ψ : L → L we write Φ ⊂ Ψ if Φ (E) ⊂ Ψ (E) for every E ∈ L. If Φ ⊂ Ψ and Φ 6= Ψ then we write Φ ⊈ Ψ.
We will consider two generalizations of Lebesgue density. First of them, called a density generated by a sequence, was introduced by J. Hejduk and M. Filipczak in [6]. For a convenience we will formulate definitions using decreasing sequences tending to zero, instead of nondecreasing sequences going to infinity.
Let e S be the family of all decreasing sequences tending to zero. We will denote sequences from e S by (a n ) or by hai. Let hai ∈ e S, E ∈ L and x ∈ ℝ. We shall say that x is an hai-density point of E (a right-hand hai-density point of E) if
n lim →∞
|E ∩ [x − a n ; x + a n ]|
2a n
= 1 ( lim
n →∞
|E ∩ [x; x + a n ]|
a n
= 1).
By Φ hai (E) (Φ + hai (E)) we will denote the set of all hai-density (right-hand hai-
density) points of E. In the same way one may define left-hand hai-density point of
E and the set Φ − hai (E). Evidently, Φ hai (E) = Φ + hai (E) ∩ Φ − hai (E).
In [6] it was proved that Φ hai is a lower density operator and the family T hai =
E ∈ L; E ⊂ Φ hai (E)
is a topology containing the density topology T d . Moreover for a n = n 1 , Φ hai = Φ d
and T hai = T d .
For any pair of sequences hai and hbi from e S we denote by hai∪hbi the decreasing sequence consisting of all elements from hai and hbi. It is clear that hai ∪ hbi ∈ e S and that
Proposition 1 If Φ hai ⊂ Φ hbi then Φ hai∪hbi = Φ hai .
The second type of density we will observe are densities generated by functions.
We denote by A the family of all functions f : (0; ∞) → (0; ∞) such that (A1) lim
x→0
+f (x) = 0, (A2) lim inf
x →0
+f(x)
x < ∞, (A3) f is nondecreasing.
Let f ∈ A. We say that x is a right-hand f-density point of a measurable set E if
h→0 lim
+|(x; x + h) \ E|
f (h) = 0.
By Φ + f (E) we denote the set of all right-hand f-density points of E. In the same way one may define left-hand f-density points of E and the set Φ − f (E). We say that x is an f -density point of E if it is a right and a left-hand f-density point of E. By Φ f (E) we denote the set of all f-density points of E, i.e. Φ f (E) = Φ + f (E) ∩ Φ − f (E).
For any f ∈ A, the family
T f = {E ∈ L; E ⊂ Φ f (E)}
forms a topology stronger than the natural topology on the real line (see [1, Th. 7]
and [4, Th. 1]).
Proposition 2 ([4, Prop. 4]) For each f, g ∈ A, an inclusion T f ⊂ T g holds if and only if Φ f ⊂ Φ g .
In [2] and [3] it has been shown that properties of f-density operator Φ f and f -density topology T f are strictly depended on lim inf
x →0
+f(x)
x . In our paper we are interested in topologies generated by functions f ∈ A for which lim inf
x→0
+f(x)
x > 0.
The family of all such functions we will denote by A 1 .
Topologies generated by functions from A 1 and from A \ A 1 have quite different
properties (for example they satisfy different separation axioms - see [3, Th. 5 and
Th. 7)]). On the other hand for any function f from A 1 properties of T f are similar to properties of topologies generated by sequences, so similar to properties of the density topology T d (compare [4, Th. 3], [9], [7] and [8]). Moreover
Proposition 3 ([4, Th. 5]) For any sequence hsi ∈ e S, the function f defined by a formula
f (x) = s n for x ∈ (s n+1 ; s n ] belongs to A 1 and Φ hsi = Φ f .
Corollary 4 For each hai , hbi ∈ e S, an inclusion T hai ⊂ T hbi holds if and only if Φ hai ⊂ Φ hbi .
In [4] we have constructed a function f ∈ A 1 such that Φ f 6= Φ hsi for each hsi ∈ e S. We will remain the definition of that function, omitting the proof, because we will construct a similar one in Theorem 11.
Example 5 ([4, Th. 6]) Let us define sequences hwi = (2, 2, 3, 3, 3, 4, 4, 4, 4, . . .) ,
hri = (1, 2, 1, 2, 3, 1, 2, 3, 4, . . .) , a 0 = 1, a n = a n −1
w 2 n for n 1, b n = a n w n = a n −1
w n for n 1.
Evidently
n lim →∞
a n −1
b n
= lim
n →∞
b n
a n = ∞.
The function f defined by a formula f (x) =
a n −1 for x ∈ (b n ; a n −1 ] , b n r n for x ∈ (a n ; b n ] . belongs to A 1 and Φ f 6= Φ hsi for each hsi ∈ e S.
The function constructed above proves that the family
T f ; f ∈ A 1 is big- ger than n
T hsi ; hsi ∈ e S o
. We will show that there exist continuum topologies from
T f ; f ∈ A 1
\ n
T hsi ; hsi ∈ e S o
and that for any pair of sequences hai , hbi ∈ e S satis- fying T hai ⊈ T hbi there is a function f ∈ A 1 such that T hai ⊈ T f ⊈ T hbi and T f 6= T hsi
for hsi ∈ e S.
Set
f α (x) = f x α
and s α (n) = αs (n)
for f ∈ A 1 , hsi ∈ e S and α > 0. Obviously f α ∈ A 1 and hs α i ∈ e S.
Proposition 6 If Φ f = Φ hsi then Φ f
α= Φ hs
αi .
Proof It is sufficient to prove that for any measurable set E 0 ∈ Φ + f
α(E) ⇐⇒ 0 ∈ Φ + hs
αi (E) . If 0 ∈ Φ + f
α(E) then
(0; x) ∩ α 1 E 0
f (x) =
1
α |(0; αx) ∩ E 0 | f α (αx)
x →0+
−→ 0,
and so 0 ∈ Φ + f 1
α E = Φ + hsi α 1 E . Hence
|(0; s α (n)) ∩ E 0 | s α (n) =
(0; s (n)) ∩ α 1 E 0 s (n)
n →∞
−→ 0,
which gives 0 ∈ Φ + hs
αi (E).
Conversely, if 0 ∈ Φ + hs
αi (E) then (0; s (n)) ∩ α 1 E 0
s (n) = | (0; s α (n)) ∩ E 0 | s α (n)
n→∞ −→ 0,
and consequently 0 ∈ Φ + hsi α 1 E = Φ + f α 1 E . Thus
|(0; x) ∩ E 0 |
f α (x) = α 0; α x
∩ α 1 E 0
f x α x→0+ −→ 0,
which implies 0 ∈ Φ + f
α(E). ■
Corollary 7 If Φ f ∈ / n
Φ hsi ; hsi ∈ e S o
then Φ f
α∈ / n
Φ hsi ; hsi ∈ e S o
for any α > 0.
Theorem 8 If f is the function defined in Example 5 then Φ f
α⊈ Φ f
βfor any α > β > 1.
Proof Since f is nondecreasing, we have f α ¬ f β and Φ f
α⊂ Φ f
β. Let (n i ) be an increasing sequence of positive integers such that r n
i= 1 for every i. Define
E = ℝ \ [ ∞ i=1
βb n
i; βb n
i√ w n
i.
It is sufficient to show that
0 ∈ Φ f
β(E) \ Φ f
α(E) .
Without loss of generality, we can assume that for any n βa n < b n and βb n < αb n < βb n √ w n < a n −1 . If x ∈ (βb n
i; βa n
i−1 ] for some i then
(1) |(0; x) ∩ E 0 |
f β (x) ¬ βb n
i√ w n
if
x β
= βb n
i√ w n
ia n
i−1
= β
√ w n
i.
If x ∈ (βa n
i; βb n
i] for some i then
(2) |(0; x) ∩ E 0 |
f β (x) < a n
if
x β
= a n
ir n
ib n
i= 1 w n
i.
If x ∈ (βa p ; βb p −1 ] and n i < p < n i+1 for some i then
(3) |(0; x) ∩ E 0 |
f β (x) < a p
r p b p ¬ 1 w p
. From (1)-(3) we conclude that 0 ∈ Φ f
β(E).
On the other hand for x i = αb n
i|(0; x i ) ∩ E 0 |
f α (x i ) |(βb n
i; αb n
i)|
f (b n
i) = (α − β) b n
ir n
ib n
i= α − β > 0,
and consequently 0 / ∈ Φ f
α(E). ■
In [4, Th. 1] it has been proved that for each function f ∈ A there is a continuous function g ∈ A such that Φ f = Φ g . Thus there is at most continuum different f- density topologies. Combining this result with Corollary 4, Corollary 7 and Theorem 8 we obtain
Corollary 9 The family
T f ; f ∈ A 1
\ n
T hsi ; hsi ∈ e S o
has cardinality continuum.
In the next theorem we will compare densities generated by sequences. Let us formulate a useful lemma from [5]. Fix hai and hbi from e S. There is an unique sequence (k n ) of positive integers such that
b n ∈ (a k
n+1 ; a k
n]
for each n with b n < a 1 . From now on (k n ) will denote this unique sequence.
Lemma 10 ([5, Th. 7]). The following conditions are equivalent (a) Φ hai ⊂ Φ hbi .
(b) For arbitrary increasing sequence (n i ) of positive integers lim inf
i→∞
a k
nib n
i< ∞ or lim inf i→∞ b n
ia k
ni+1 = 1.
Theorem 11 Let hai , hbi ∈ e S. If Φ hbi ⊈ Φ hai then there is f ∈ A 1 such that Φ hbi ⊂ Φ f ⊂ Φ hai and Φ f 6= Φ hsi for hsi ∈ e S.
Proof By Proposition 1 we can asume that hai is a subsequence of hbi. Since Φ hai ⊈ Φ hbi , there exists an increasing sequence (n i ) of positive integers such that
i lim →∞
a k
nib n
i= ∞ and M = lim inf i
→∞
b n
ia k
ni+1 > 1.
Let us denote
β =
√
4M ; M < ∞
2 ; M = ∞ .
Replacing (n i ) by a subsequence we can assume that (k n
i) is increasing and for every i
(4) a b
knini
> i 2 , a b
nikni +1
> β 3 and a b
nikni +1
< β 5 if M < ∞.
Let us define
hci = hai ∪ (b n
i) ,
hri = (1, 2, 1, 2, 3, 1, 2, 3, 4, . . .) , g 1 (x) = a n for x ∈ (a n+1 ; a n ] , g 2 (x) = c n for x ∈ (c n+1 ; c n ] , g 3 (x) = b n for x ∈ (b n+1 ; b n ] ,
f (x) =
r i b n
ifor x ∈ a k
ni+1 ; b n
i, g 1 (x) for x / ∈ S ∞
i=1 a k
ni+1 ; b n
i.
Since 1 ¬ r i ¬ i 2 , f ∈ A 1 and g 3 ¬ g 2 ¬ f ¬ g 1 . Thus
Φ hbi = Φ g
3⊂ Φ hci = Φ g
2⊂ Φ f ⊂ Φ g
1= Φ hai .
We will show that Φ f 6= Φ hsi for hsi ∈ e S. Suppose, contrary to our claim, that there is hsi ∈ e S such that
(5) Φ f = Φ hsi .
Let us consider the sets
T =
i ∈ ℕ; ∃ m ∈ℕ s m ∈
β 3 a k
ni+1 ; ib n
i, R = {r i ; i ∈ T } .
We will prove that the set R is bounded. Suppose it is not true and so T is infinite.
Let (t i ) be an increasing sequence consisting of all elements from T , i.e.
T = {t i ; i ∈ ℕ} .
Let m i be a fixed positive integer such that s m
i∈ h
β 3 a k
nti+1 ; t i b n
tii .
We will show that
(6) s m
i< 2b n
tifor almost every i. Suppose on the contrary that there is an increasing sequence (p i ) of positive integers such s m
pi 2b n
tpifor every i. Hence
a k
ntpi
s m
pi t 2 p
ib n
tpit p
ib n
tpi= t p
i p i i
and so
i lim →∞
a k
ntpi
s
mpi= ∞ and
lim inf
i →∞
s
mpib n
tpi 2.
This, by Lemma 10, contradicts the assumption that Φ hci ⊂ Φ hsi and finishes the proof of (6).
Now we show that there are real numbers ε > δ > 0 such that (7) a k
nti+1 < δb n
ti< εb n
ti< s m
ifor almost every i. If M = lim inf
i →∞
b
nia
kni +1< ∞ then by (4) it is sufficient to set ε = β 1
2and δ = β 1
3. If M = ∞, then obviously it is sufficient to show that for some positive ε inequality
εb n
ti< s m
iholds for almost every i. Suppose on the contrary that there is an increasing sequence (p i ) of positive integers such is m
pi< b n
tpifor every i. Hence
i lim →∞
b n
tpis m
pi= ∞.
and
lim inf
i →∞
s m
pia k
ntpi
+1 β 3 > 1.
which also contradicts the assumption that Φ hci ⊂ Φ hsi and ends the proof of (7).
By our assumptions, the set R is not bounded. Thus there exists an increasing sequence (p i ) of positive integers such that
i→∞ lim r t
pi= ∞.
Put
A = [ ∞ i=1
δb n
tpi; εb n
tpi.
From (6) and (7) it follows that A ∩ 0; s m
pis m
pi εb n
tpi− δb n
tpi2b n
tpi= ε − δ 2 > 0 for almost every i, and so
(8) 0 / ∈ Φ + hsi (A 0 ).
We will show that 0 ∈ Φ + f (A 0 ). Let x ∈ (a j ; a j −1 ]. If j = k n
tpifor some i then
(9) |A ∩ (0; x)|
f (x) ¬ εb n
tpir t
pib n
tpi= ε r t
pi.
On the other hand if k m
ni< j < k m
ni+1for some i then
(10) |A ∩ (0; x)|
f (x) ¬ εb n
tpi+1a j+1 ¬ εb n
tpi+1a k
ntpi+1